# What is the concentration of a KOH (aq) solution if 12.8 mL of this solution is required to react with 25.0 mL of .110 mol/L H_2SO_4(aq)?

Jan 17, 2016

We assume stoichiometric quantities of potassium hydroxide and sulfuric acid.

#### Explanation:

$2 K O H \left(a q\right) + {H}_{2} S {O}_{4} \rightarrow {K}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

The reaction represents the neutralization of sulfuric acid by potassium hydroxide.

Moles of sulfuric acid: $25.0 \times {10}^{-} 3 \cancel{L} \times 0.110 \cdot m o l \cdot \cancel{{L}^{-} 1}$ $=$ ?? $m o l$

From the equation above, which shows that the stoichiometry of hydroxide to sulfuric acid is $2 : 1$, and the volume of stoichiometric acid:

$\left[K O H\right] = \frac{25.0 \times {10}^{-} 3 \cancel{L} \times 0.110 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 2}{12.8 \times {10}^{-} 3 L}$ $=$ ??mol*L^-1.

We get an answer in $m o l \cdot {L}^{-} 1$, as required.